Question: $\begin{aligned} &H(x)=3x^2-6 \\\\ &h(x)=H'(x) \end{aligned}$ $\int_{-1}^{0} h(x)\,dx=$
Solution: $h$ is the derivative of $H$, which means $H$ is an antiderivative of $h$. Since we know the antiderivative of $h$, we can use the fundamental theorem of calculus: For every function $h$ and its antiderivative $H$, $\int_a^b h(x)\,dx=H(b)-H(a)$. $\begin{aligned} &\phantom{=}\int_{-1}^{0} h(x)\,dx \\\\ &=H({0})-H({-1}) \\\\ &=[3{(0)}^2-6]-[3{(-1)}^2-6] \\\\ &=-6-(-3) \\\\ &=-3 \end{aligned}$ In conclusion, $\int_{-1}^{0} h(x)\,dx=-3$